14. LECTURE 14. CELESTIAL MECHANICS 151

14 Lecture 14. Celestial mechanics

14.1 Two-body summaryEverybody should be very familiar with the two celestial body dynamics. Here Imention some topics that may not appear in elementary expositions.

⟨⟨Bertrand’s theorem⟩⟩ Suppose 𝑈 is a spherically symmetric potential. For amotion with non-zero angular momentum close to a circle to have a closed orbit, 𝑈must be harmonic or gravitational.140

⟨⟨Collision and extension⟩⟩ If the angular momentum is zero, the particles cancollide. The orbit may be, however, uniquely extended beyond collision.141

14.2 Necessary condition for stabilityThe general 𝑛-body problem consists of 𝑛 point masses (𝑚1, 𝑟1), · · ·, (𝑚𝑛, 𝑟𝑛) at-tracting one another according to the law of gravity. The total kinetic energy is

𝐾 =1

2

∑︁𝑖

𝑚𝑖�̇�2𝑖 , (14.1)

and the potential energy 𝑈 is

𝑈 = −∑︁𝑖<𝑗

𝑚𝑖𝑚𝑗

|𝑟𝑖 − 𝑟𝑗|. (14.2)

We describe the system from the inertial frame and the origin of the position coor-dinates is the center of mass.

We say the system is stable, if(a) No collision: |𝑟𝑖 − 𝑟𝑗| > 0 for all 𝑖 and 𝑗 for all 𝑡.(b) No escape: there is 𝑐 > 0 such that |𝑟𝑖| < 𝑐 for all 𝑖 and 𝑡.

The fate of a three body system is very hard to predict (Fig. 14.1)

14.3 Jacobi’s necessary condition for non-escape142

If there is no escape nor collisions, the total energy of the system must be negative.

140S. A. Chin, A truly elementary proof of Bertrand’s theorem, arXiv:1411.7057 [physics.class-ph](2014).

141Arnold III p56142Arnold III p59

152

t = 0, 1, ... , 10

t = 40, 41, ... , 50= 50, 41, ... , 60t = 60, 41, ... , 70t

Figure 14.1: Fate of a three body system [Fig. 12-15 of Arnold III ]

Remark For 𝑛 > 2, this is not sufficient as seen in Fig. 14.1.[Demo]We use Lagrange’s formula:*143 for 𝐼 =

∑︀𝑚𝑖𝑟

2𝑖 (the moment of inertia) and the total

energy 𝐸𝐼 = 4𝐸 − 2𝑈. (14.3)

Since 𝑈 < 0, 𝐸 ≥ 0 implies 𝐼 > 0, so 𝐼(𝑡) must be convex. Therefore, it cannot be

143

𝐼 =∑︁𝑖

𝑚𝑖(2�̇�2𝑖 + 2𝑟𝑖𝑟𝑖) = 4𝑇 − 2

∑︁𝑖

𝑟𝑖∑︁𝑗 ̸=𝑖

𝑚𝑖𝑚𝑗

|𝑟𝑖 − 𝑟𝑗 |3(𝑟𝑖 − 𝑟𝑗)

(Note that ∇(1/|𝑥|) = −(1/|𝑥|2)(𝑥/|𝑥|), because ∇|𝑥|2 = 2|𝑥|∇|𝑥| = 2𝑥). Therefore,

𝐼 = 4𝑇 −∑︁𝑖

𝑟𝑖∑︁𝑗

𝑚𝑖𝑚𝑗

|𝑟𝑖 − 𝑟𝑗 |3(𝑟𝑖 − 𝑟𝑗)−

∑︁𝑗

𝑟𝑗∑︁𝑖

𝑚𝑖𝑚𝑗

|𝑟𝑖 − 𝑟𝑗 |3(𝑟𝑗 − 𝑟𝑖)

= 4𝑇 − 2∑︁𝑖,𝑗

𝑚𝑖𝑚𝑗

|𝑟𝑖 − 𝑟𝑗 |3(𝑟𝑖 − 𝑟𝑗)

2 = 4𝑇 + 2𝑈 = 4𝐸 − 2𝑈.

14. LECTURE 14. CELESTIAL MECHANICS 153

bounded from above. Note that*144

𝐼∑︁𝑖

𝑚𝑖 =∑︁𝑖<𝑗

𝑚𝑖𝑚𝑗|𝑟𝑖 − 𝑟𝑗|2 +

(︃∑︁𝑖

𝑚𝑖𝑟𝑖

)︃2

. (14.4)

The center of mass is fixed (say, 0), so the unbounded 𝐼 means the increase of themutual distance without bound.

If there is no collision and 𝑈 is bounded from below, the virial theorem145 may beapplied to the time average of 𝑈 : ⟨𝑈⟩ = 2𝐸. Therefore, 𝐸 < 0.

14.4 Collisions146

If there is a simultaneous 𝑛-body collision 𝐼 vanishes. If 𝐼(𝑡) → 0 as 𝑡→ 𝑡0, the thetotal angular momentum must be zero.

For binary collisions the motion can be smoothly extended beyond collisions. Thismeans that for a three body problem with nonzero angular momentum, the motionis well defined for all 𝑡.147

14.5 Three-body problem and its reductionFor the 𝑛 = 3 case (the three-body problem case) the original equation of motion is3 × 3 × 2 = 18 first order equations. The problem can be reduced to a problem of 6first order differential equations (Lagrange 1772).(i) The location and the velocity of the center of mass are cyclic coordinates (= thecoordinate that do not explicitly appear in the equations), so we may regard them

144

𝐼∑︁𝑖

𝑚𝑖 =∑︁𝑖,𝑗

𝑚𝑖𝑚𝑗𝑟2𝑖 =

1

2

∑︁𝑖,𝑗

𝑚𝑖𝑚𝑗(𝑟2𝑖 + 𝑟2𝑗 ) =

1

2

∑︁𝑖,𝑗

𝑚𝑖𝑚𝑗 [(𝑟𝑖 − 𝑟𝑗)2 + 2𝑟𝑖𝑟𝑗 ].

145 ⟨⟨Virial theorem⟩⟩ Notice that the long-time average of the derivative of a bounded function

f(t) vanishes: (1/𝑇 )∫︀ 𝑇

0𝑓 ′(𝑠)𝑑𝑠 → 0. Consider 𝑝𝑞. If the phase space for the system is bounded,

then the time average of 𝑑(𝑝𝑞)/𝑑𝑡 vanishes. Therefore, for 𝑈 which is a homogenous function ofdegree −1

𝑝𝑞 + 𝑞�̇� = 2𝑇 − 𝑞𝜕𝑈

𝜕𝑞= 2𝑇 + 𝑈 = 2𝐸 − 𝑈.

Therefore, ⟨𝑈⟩ = 2𝐸.146Arnold III p59147details can be found on p60-61 of Arnold III.

154

to be known. Thus 12 equations remain.(ii) The total angular momentum is conserved. Thus 9 equations remain.(iii) We perform the elimination of the nodes. The rotation as a whole may be re-garded as known; actually this is done with (ii). Thus, 8 equations remain.(iv) The total energy is conserved: 7 equations now.(v) Now, eliminate 𝑡 from the equation. The resultant set contains 6 first orderequations.

The actual procedures and formulas are explained in detail on p343-347 of Whit-taker.

14.6 Restricted problem of three bodyIf the third body (planetoid P) has an infinitesimal mass moving in the plane of themotion of the other two bodies S (Sun) and J (Jupiter) under their influence, thethree-body problem is called the restricted problem of three bodies. The Hamiltoniangoverning the motion of P is given by

𝐻 =1

2(𝑈2 + 𝑉 2) − 𝑚1

SP− 𝑚2

JP. (14.5)

SP ad JP are time-dependent, so 𝐻 is not a constant of motion.We introduce a moving coordinate system Fig. 14.2 whose origin is the CM of S

and J and whose 𝑥-axis is from O to J. 𝑂-𝑥 is chosen to span the plane on which Sand J always sit. Let 𝑛 be the angular speed of SJ.

o

J

xy

Figure 14.2: Moving coordinates for the restricted problem

Then, the original position coordinates (𝑋, 𝑌 ) of P may be related to (𝑥, 𝑦) as

𝑋 = 𝑥 cos𝑛𝑡− 𝑦 sin𝑛𝑡 (14.6)

𝑌 = 𝑥 sin𝑛𝑡+ 𝑦 cos𝑛𝑡. (14.7)

Now, we wish to canonical transform the coordinate system from the inertial (𝑋, 𝑌 )(the old (𝑞, 𝑝)) to the moving (𝑥, 𝑦) (the new (𝑄,𝑃 )). We use 𝑊 = 𝑝𝑞−𝐹 , where 𝐹

14. LECTURE 14. CELESTIAL MECHANICS 155

is the ‘most’ standard one introduced in 13.1

𝑑𝑊 = 𝑃𝑑𝑄−𝐾𝑑𝑡+ 𝑞𝑑𝑝+𝐻𝑑𝑡 = 𝑢𝑑𝑥+ 𝑣𝑑𝑦 +𝑋𝑑𝑈 + 𝑌 𝑑𝑉 + (𝐻 −𝐾)𝑑𝑡. (14.8)

Since 𝑑𝑊 is exact, we can integrate this as 𝑊 = 𝑝𝑞 (i.e., in terms of the old variables).Expressing 𝑞 in terms of the new coordinates, we get

𝑊 = 𝑈(𝑥 cos𝑛𝑡− 𝑦 sin𝑛𝑡) + 𝑉 (𝑥 sin𝑛𝑡+ 𝑦 cos𝑛𝑡). (14.9)

Note that

𝑋 =𝜕𝑊

𝜕𝑈, 𝑌 =

𝜕𝑊

𝜕𝑉, 𝑢 =

𝜕𝑊

𝜕𝑥, 𝑣 =

𝜕𝑊

𝜕𝑦. (14.10)

Since our new coordinates are time-dependent, the new Hamiltonian ahas an extraterm:*148

𝜕𝑊

𝜕𝑡= 𝐻 −𝐾, or 𝐾 = 𝐻 − 𝜕𝑊

𝜕𝑡=

1

2(𝑢2 + 𝑣2) + 𝑛(𝑢𝑦 − 𝑣𝑥) − 𝐹, (14.11)

where𝐹 =

𝑚1

SP+𝑚2

JP(14.12)

is a function of 𝑥 and 𝑦 only, so note that 𝐾 is time independent. 𝐾 = const is anintegral and called the Jacobian integral. If the sum of the masses of S and J to bechosen unity, we may rewrite 𝐹 as

𝐹 =1 − 𝜇

SP+

𝜇

JP. (14.13)

If 𝜇 = 0, we ignore the effect of J. Thus, 𝜇 could be regarded as a perturbationparameter.

A nice restricted three-body simulation video is:https://www.youtube.com/watch?v=jarcgP1rRWs

148

𝜕𝑊

𝜕𝑡= 𝑈(−𝑛𝑥 sin𝑛𝑡− 𝑛𝑦 cos𝑛𝑡) + 𝑉 (𝑛𝑥 cos𝑛𝑡− 𝑛𝑦 sin𝑛𝑡).

Also we have

𝑢 =𝜕𝑊

𝜕𝑥= 𝑈 cos𝑛𝑡+ 𝑉 sin𝑛𝑡, 𝑣 =

𝜕𝑊

𝜕𝑦= −𝑈 sin𝑛𝑡+ 𝑉 cos𝑛𝑡. (*)

Therefore,

𝜕𝑊

𝜕𝑡= 𝑥(−𝑛𝑈 sin𝑛𝑡+ 𝑛𝑉 cos𝑛𝑡) + 𝑦(−𝑛𝑈 cos𝑛𝑡− 𝑛𝑉 sin𝑛𝑡) = 𝑛(𝑥𝑣 − 𝑦𝑢).

Obviously, 𝑈2 + 𝑉 2 = 𝑢2 + 𝑣2, since (*) is an orthogonal transformation.

156

14.7 Bruns’ theoremTheorem [Bruns 1887] The classical integrals (energy, momentum and angular mo-mentum) are the only independent algebraic integrals of the problem of three bodies.

A proof can be found on p359-377 of Whittaker. The proof fully utilizes the pe-culiarity of the three-body problem.

14.8 Poincare’s theorem ‘denying’ integrability of perturbed systemsSuppose we have a completely integrable system, whose Hamiltonian is given interms of action variables only as 𝐻0(𝐼) (with non-resonance condition satisfied, i.e.,the Hessian of 𝐻0(𝐼) is non-singular). The perturbation term 𝐻1(𝜃, 𝐼) is analytic in𝐼 and the angle variable 𝜃. The total Hamiltonian reads

𝐻(𝜃, 𝐼, 𝜇) = 𝐻0(𝐼) + 𝜇𝐻1(𝜃, 𝐼). (14.14)

Then, under a natural condition, there is no first integral of motion other than 𝐻itself that are analytic in 𝜇.

Poincare proved in 1889 this theorem for the system with two degrees of freedom(as in the restricted three-body problem).

Suppose there is a first integral of motion Φ

Φ = Φ0 + 𝜇Φ1 + 𝜇2Φ2 + · · · . (14.15)

The key observation is that Φ0 must be a function of 𝐻0, which follows from theobservation that Φ0 is a function of 𝐼 only. If Φ depends on 𝜇 smoothly, perhapsit is not a wild guess that this structure must be preserved for Φ (i.e., Φ must be afunction of 𝐻 only). A proof is given in 14.10-14.12.149

14.9 Significance of Poincare’s negative resultBruns proved150 in 1887 that for the three-body problem algebraic integrals of motionare exhausted by the classical integrals of motion (3 CM coordinates− 𝑡 × veloc-ity, 3 angular momentum components, 3 momentum components and energy), so nosimplification further than accomplished by Lagrange long ago (see 14.5). Then,

149based on H. Yoshida 5.1 in Y. Ohnuki and H. Yoshida, Mechanics (Iwanami 1994).150E. T. Whittaker, A treatise on the analytical dynamics of particles and rigid bodies (Cambridge,

1937 [4th edition]) p358-377.

14. LECTURE 14. CELESTIAL MECHANICS 157

Poincare proved even for the restricted simpler problem there is no generally smooth(wrt 𝜇) integration possible. Thus, almost all the people lost hope in solving thecelestial mechanics in a closed form.

However, do not forget that Poincare did not show that for fixed values of 𝜇 ana-lytic Φ exists (a less smooth Φ may exist).

14.10 Φ0 is a function of 𝐼 onlyLet us Fourier-expand Φ0:

Φ0(𝜃, 𝐼) =∑︁𝑘∈Z2

𝜑𝑘(𝐼)𝑒𝑖𝑘·𝜃. (14.16)

The Fourier expansion of [𝐻0(𝐼),Φ0]𝑃𝐵 = 0 reads

𝜕

𝜕𝐼𝐻0(𝐼) · 𝜕

𝜕𝜃Φ0 = ∇𝐻0(𝐼) ·

∑︁𝑘∈Z2

(𝑖𝑘)𝜑𝑘(𝐼)𝑒𝑖𝑘·𝜃 = 0. (14.17)

Therefore, for all 𝑘∇𝐻0(𝐼) · 𝑘𝜑𝑘(𝐼) = 0. (14.18)

This must be true for any 𝐼, so its derivative wrt 𝐼 must vanish. Thus,

Hess(𝐻0(𝐼))𝑘𝜑𝑘(𝐼) + ∇𝐻0(𝐼) · 𝑘∇𝜑𝑘(𝐼) = 0. (14.19)

If 𝜑𝑘(𝐼) ̸= 0, then (14.18) implies ∇𝐻0(𝐼) ·𝑘 = 0, so Hess(𝐻0(𝐼))𝑘 must vanish.But since the Hessian is non-singular, this is true only for 𝑘 = 0. Thus, only𝜑0(𝐼) ̸= 0. That is, Φ0 cannot depend on 𝜃.

14.11 Φ0 is a function of 𝐻0 onlyThis is demonstrated under a complicated condition whose significance is un-clear to me. So let us assume this. That is (recall that our system has 2 degreesof freedom),

𝜕(𝐻0,Φ0)

𝜕(𝐼1, 𝐼2)= 0. (14.20)

This implies151 the existence of a function 𝜓 such that Φ0 = 𝜓 ∘𝐻0.

151This conclusion follows from (14.20) if everything is smooth enough; of course, we have assumeeverything is holomorphic, so the argument is OK. Since Jacobian is the volume ratio of 𝑑𝐼1𝑑𝐼2and 𝑑𝐻0𝑑Φ0, (14.20) implies 𝑑𝐻0 and 𝑑Φ0 are parallel.

158

14.12 Conclusion of proof: Φ is a function of 𝐻 onlyTake 𝜓 in 14.11 and make Φ − 𝜓(𝐻). This is Φ0 − 𝜓(𝐻0) = 0 for 𝜇 = 0.Therefore, we may write

Φ(𝜃, 𝐼;𝜇) − 𝜓(𝐻(𝜃, 𝐼;𝜇)) = 𝜇Φ(1)(𝜃, 𝐼;𝜇), (14.21)

which is an integral of motion as well. Expanding this,

Φ(1) = Φ(1)0 + 𝜇Φ

(1)1 + · · · , (14.22)

we see that Φ(1)0 is a function of 𝐻0 (a function of 𝐼 only is a function of 𝐻0):

Φ(1)0 = 𝜓(1)(𝐻0), (14.23)

so we can repeat that argument as

Φ(1) − 𝜓(1)(𝐻) = 𝜇Φ(2)(𝜃, 𝐼;𝜇). (14.24)

Thus,

Φ = 𝜓(𝐻) + 𝜇Φ(1) (14.25)

= 𝜓(𝐻) + 𝜇(𝜓(1)(𝐻) + 𝜇Φ(2)) (14.26)

= 𝜓(𝐻) + 𝜇𝜓(1)(𝐻) + 𝜇2Φ(2) (14.27)

· · · (14.28)

= 𝜓(𝐻) + 𝜇𝜓(1)(𝐻) + 𝜇2𝜓(2)(𝐻) + · · · (14.29)

This implies that if Φ can be obtained perturbatively, it is not an independentinvariant of motion.

14.13 Asteroids and gapsDue to Jupiter they could not form a planet (inside the so-called frost line). Thetotal mass is about 4% of Moon and 40% of the total mass is concentrated in Ceresand Vesta.Introductory video (this is for kids, but good)

https://www.youtube.com/watch?v=iy19nHTVLEY

All known asteroidshttps://www.youtube.com/watch?v=vfvo-Ujb_qk

There are several major gaps in the distribution of asteroids called Kirkwood gaps(discovered 1866 by D Kirkwood 1814-1895). He correctly explained their origin interms of motional resonance with Jupiter (e.g., 3:1 at 2.5 AU; see Fig. 14.3).

14. LECTURE 14. CELESTIAL MECHANICS 159

Figure 14.3: Asteroid belt and Kirkwood gaps

14.14 Special solutions of restricted three-body problemNeedless to say, many people tried to find special solutions (orbits) for the restrictedthree-body problem. There are 5 equilibrium points in the moving coordinate system(𝑥, 𝑦). Three of them are along the SJ axis (Euler’s linear solutions), but they arenot stable. The remaining two are Lagrange’s regular triangle solutions: the pointsare at the apexes of the regular triangles on the rotational plane (the plane spannedby (𝑥, 𝑦)) whose one edge is the SJ segment.

Are the Lagrange points stable? For this to be stable it is not very hard to showthat 𝜇(1 − 𝜇) < 1/27 (or 𝜇 < 0.0385 · · ·). To show the stability actually, we mustshow the existence of small invariant tori surrounding these orbits. Thus, it is relatedto KAM and highly nontrivial, but except for three values of 𝜇 < 0.0385 · · · theirstability was shown by Arnold.

Figure 14.4: A gravitational potential contour plot showing Earth’s Lagrangian points; L4 andL5 are above and below the planet, respectively. [Wikipedia: Jupiter Trojan]

160

The equilibrium points may not be so hard to find from the gravitational potentialplot (Fig. 14.4, although this is for the earth). Lagrange’s points are potential maxpoints. Then, how can a particle stay near the hill tops? Physically speaking, thisis due to Coriolis’ force.

We actually observe asteroids around Lagrange’s equilibrium points along Jupitersorbit (the so-called Trojan and Greek asteroid groups; the Trojan ‘camp’ trailsJupiter). The Hilda group asteroids are strong resonance with Jupiter. A goodillustration of Trojan (green), Greek (red), and Hilda (white) families of asteroids is:

https://people.duke.edu/~ng46/borland/hilda%20family.gif.152

Here the Hilda is separated:https://www.youtube.com/watch?v=yt1qPCiOq-8

14.15 Did asteroids cause mass extinctions?The biggest mass extinction (the end Permian MS) is certainly not due to asteroids.I am not so convinced by the physicists’ asteroid theory of the KPg mass extinction(65.5 MaBP; Cretaceous-Paleogene mass extinction ‘due to’ an impact at Chicxu-lub). It is sure that an asteroid hit had a severe effect. However, this does not meanthe asteroid hit was the main cause; it could well be the last straw. Remember thatbig scale mass extinctions are always with drastic sea level changes as seen in Fig.14.5. How can asteroid hit be predicted by the sea level changes? It should be fair toclaim that an asteroid can cause havoc only when the biosphere is strained severelyalready.

Figure 14.5: Mass extinctions and see level changes

152from DrBill’s Astronomy Web Site http://hildaandtrojanasteroids.net

14. LECTURE 14. CELESTIAL MECHANICS 161

14.16 RingsSaturn detail:

https://www.youtube.com/watch?v=ENwQ7-qLlrA